Did the result of the Hilbert hotel paradox change after the proof $\mathfrak p=\mathfrak t$?

Question

We have seen questions like What is the result of $\infty-\infty$ ? in 2011 and the result was that it is indeterminate. I find the example of $\infty-\infty=7$ absolutely convincing.

We now (2016; I'm late to the party) have the proof that $\mathfrak p=\mathfrak t$, i.e. that different infinities are actually the same. Does that change the Hilbert hotel result from $\infty-\infty=7$ (or anything) to the single result $\infty-\infty=0$?

If possible, please explain in simple words like the linked answer, which was perfect for me.

Answer 1

No, $\mathfrak{p}$ and $\mathfrak{t}$ are two specific infinities that are in some sense (possibly) bigger than the infinity denoted as $\infty$ in that question.

Your question makes as much sense as: "I had two different bags of stones, and I checked, and it turns out that the two bags actually contain the same number of stones. Does that mean anything about this third bag of stones I just found?". I believe this analogy is deceptively good, since $\mathfrak{p}$ and $\mathfrak{t}$ are cardinals, which are just a generalisation of what we normally mean by "number of things in a thing"; two sets have the same cardinality if there is some way to match up the elements one-for-one without missing any elements on either side.

I strongly suspect that your mental model is missing a very large piece, namely the fact that there are (infinitely!) many different cardinals: $\mathbb{N}$ is strictly smaller than $\mathbb{R}$ (look up Cantor's diagonal argument, for example).

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I'll actually digress and give my favourite argument that $\mathbb{R}$ is uncountable, rendered from http://people.math.gatech.edu/~mbaker/pdf/realgame.pdf:

Alice and Bob play a game. Alice starts at 0, Bob starts at 1, and they alternate taking turns (starting with Alice), each picking a number between Alice and Bob’s current numbers. (So start with $A:0$, $B:1$, then $A:0.5$, $B:0.75$, $A:0.6$, $\dots$ is an example of the start of a valid sequence of plays.) We fix a subset $S$ of $[0,1]$ in advance, and Alice will win if at the end of all time the sequence of numbers she has picked converges to a number in $S$; Bob wins otherwise. (Alice’s sequence does converge: it’s increasing and bounded above by $1$.)

It’s obvious that if $S = [0,1]$ then Alice wins no matter what strategy either of them uses: a convergent sequence drawn from $[0,1]$ must converge to something in $[0,1]$.

Also, if $S = \{s_1, s_2, \dots\}$ can be matched up one-to-one with $\mathbb{N}$ (that is, $S$ is countable, where I just wrote down a matching of $S$ with $\mathbb{N}$) then Bob has a winning strategy: at move $n$, pick $s_n$ if possible, and otherwise make any legal move. (Think for a couple of minutes to see why this is true: if Bob couldn’t pick $s_n$ at time $n$, then either Alice has already picked a number bigger, in which case she can’t ever get back down near $s_n$ again, or Bob has already picked a number $b$ which is smaller, in which case she is blocked off from reaching $s_n$ because she can’t get past $b$.)

So if $[0,1]$ is countable then Alice must win no matter what either of them does, but Bob has a winning strategy; contradiction.

Answer 2

Nothing has changed about the Hilbert hotel. It concerns countable infinite sets, that is, infinite sets $N$ such that there is a bijection between $N$ and the set $\Bbb N$ of natural numbers. The equality $\mathfrak p=\mathfrak t$ has nothing to do with this.

But, in fact, if $I$ is an infinite set and if $f$ is a finite subset of $I$ with, say $7$ elements, then $I\setminus F$ has the same cardinal as $I$ and $I\setminus(I\setminus F)=F$. In this is, the equality $\infty-\infty=7$ still holds.

Answer 3

There are different types of "sizes" that a collection of items can have. Three main ones are ordinality, cardinality, and measure. The basis of the Hilbert Hotel is that different ordinalities have the same cardinality (two collections can have the same cardinality but different ordinality, but if they have the same ordinality, they have the same cardinality). I said "collection" rather than "set" because we have ordinalities only when order matters. For a set, where the order doesn't matter, we only have cardinality.

The original ordinality of the hotel is $\omega$. We can think of moving the person in Room 1 to Room 2, the person in Room 2 to Room 3, etc. as also being $\omega$. If we then have an additional person arrive after that, then that's $\omega+1$, which is a different ordinality from $\omega$. But since it's the same cardinality, we can still fit everyone in. That is, "take an infinite number of people" and "take an infinite number of people, then seven more" give the "same" number of people from a cardinality point of view, because they're just in a different order. But they're different ordinalities, because, as the answer to the other question points out, when we remove the first from the second, we have seven left over.

If we have an infinite number of groups, each group containing an infinite number of people, that has a cardinality of $\omega*\omega$ or $\omega^2$. This has the same cardinality as $\omega$, so there is a strategy for getting them into the hotel. There's also a strategy for $\omega^3$, and another for $\omega^4$, etc. For any particular finite $n$, there is some strategy to get $\omega^n$ people into the hotel. However, there's no one strategy that works for every $n$, because the set of all $\omega^n$ has a cardinality larger than $\omega$ (that is, $\omega+\omega^2+\omega^3+...$ has a larger cardinality than $\omega$).

We can get $\infty-\infty=7$ by taking the first $\infty$ to be $\omega+7$ and the second one to be $\omega$. When we "subtract" them (subtraction isn't really defined, so I'm speaking loosely here), we get $7$. If $\infty$ is understood to be a cardinality, then since we can take a bunch of different ordinalities for a particular cardinality, we have wide choice as what the "difference" of $\infty-\infty$ is. If $\infty$ is understood to be an ordinality, however, then the difference is either a particular number, or incoherent.

The cardinality of $\mathbb R$ is larger than the cardinality of $\mathbb N$, so there's no way to arrange the elements of $\mathbb R$ such that we can "subtract" $\mathbb N$ and get $0$.